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Magazine Chapter 7: Problem 55
Evaluate the integrals. $$\int_{-\pi / 4}^{\pi / 4} \cosh (\tan \theta) \sec ^{2} \theta d \theta$$
Short Answer
Expert verified
The integral evaluates to \( 2 \sinh(1) \).
Step by step solution
01
Identify the Structure of the Integral
The integral to evaluate is \( \int_{-\pi/4}^{\pi/4} \cosh(\tan \theta) \sec^2 \theta \, d\theta \). This integral involves a hyperbolic cosine function of \( \tan \theta \) and a \( \sec^2 \theta \) term, which suggests a potential substitution based on the presence of \( \sec^2 \theta \).
02
Substitution
Use the substitution \( u = \tan \theta \), hence \( du = \sec^2 \theta \, d\theta \). This substitution transforms the integral into \( \int \cosh(u) \, du \). The limits of integration need to be changed according to \( u \): when \( \theta = -\pi/4 \), \( u = \tan(-\pi/4) = -1 \), and when \( \theta = \pi/4 \), \( u = \tan(\pi/4) = 1 \).
03
Evaluate the New Integral
The integral becomes \( \int_{-1}^{1} \cosh(u) \, du \). The function \( \cosh(u) \) has the antiderivative \( \sinh(u) \). Therefore, evaluate it as \( \left[ \sinh(u) \right]_{-1}^{1} \).
04
Compute the Definite Integral
Substitute the limits into the antiderivative: \( \sinh(1) - \sinh(-1) \). Since \( \sinh \) is an odd function, \( \sinh(-u) = -\sinh(u) \). Thus, \( \sinh(-1) = -\sinh(1) \), resulting in \( \sinh(1) - (-\sinh(1)) = 2 \sinh(1) \).
05
Final Simplification
Thus, the final result of the integral is \( 2 \sinh(1) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hyperbolic Functions
Hyperbolic functions are similar to trigonometric functions but are based on hyperbolas rather than circles.
These functions, like their trigonometric counterparts, are useful in many areas of mathematics, including calculus.
In our exercise, the function involved is the hyperbolic cosine, written as \( \cosh(x) \). This function is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \), where \( e \) is the base of the natural logarithms.
These functions, like their trigonometric counterparts, are useful in many areas of mathematics, including calculus.
In our exercise, the function involved is the hyperbolic cosine, written as \( \cosh(x) \). This function is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \), where \( e \) is the base of the natural logarithms.
- The nature of the hyperbolic cosine function makes it an even function, which means \( \cosh(-x) = \cosh(x) \).
- One primary characteristic of hyperbolic functions is that they come in pairs, just like sine and cosine do in trigonometry. For \( \cosh(x) \), its pair is \( \sinh(x) \), which satisfies \( \cosh^2(x) - \sinh^2(x) = 1 \).
Definite Integrals
Definite integrals are a fundamental aspect of calculus, providing a way to compute the area under a curve.
In general, a definite integral of a function \( f(x) \) from \( a \) to \( b \) is represented as \( \int_{a}^{b} f(x) \, dx \). It is evaluated using the antiderivative \( F(x) \) of \( f(x) \): \( F(b) - F(a) \).
In our problem, we are evaluating the definite integral \( \int_{-\pi/4}^{\pi/4} \cosh(\tan \theta) \sec^2 \theta \, d\theta \).
In general, a definite integral of a function \( f(x) \) from \( a \) to \( b \) is represented as \( \int_{a}^{b} f(x) \, dx \). It is evaluated using the antiderivative \( F(x) \) of \( f(x) \): \( F(b) - F(a) \).
In our problem, we are evaluating the definite integral \( \int_{-\pi/4}^{\pi/4} \cosh(\tan \theta) \sec^2 \theta \, d\theta \).
- First, a substitution is used to simplify the integral. Here, \( u = \tan \theta \), resulting in \( du = \sec^2 \theta \, d\theta \).
- This substitution changes the limits of integration as well. Originally \( \theta \) ranged from \(-\pi/4\) to \(\pi/4\), translating to \( u \) ranging from \(-1\) to \(1\).
Trigonometric Substitution
Trigonometric substitution is an invaluable technique in integral calculus, allowing complicated integrals to be solved by replacing expressions with trigonometric identities.
This method is especially useful when dealing with functions involving square roots or quadratic expressions.
For the exercise, a trigonometric substitution was used, substituting \( u = \tan \theta \). Here鈥檚 why it works:
This method is especially useful when dealing with functions involving square roots or quadratic expressions.
For the exercise, a trigonometric substitution was used, substituting \( u = \tan \theta \). Here鈥檚 why it works:
- \( \tan \theta \) and \( \sec^2 \theta \) are related through the derivative: \( d(\tan \theta) = \sec^2 \theta \, d\theta \).
- Through substitution, the integral becomes \( \int \cosh(u) \, du \), which is far simpler to evaluate.
Odd Functions
Odd functions have a unique symmetry: they are symmetric about the origin, meaning \( f(-x) = -f(x) \).
These characteristics are useful when evaluating integrals, especially definite integrals over symmetric limits like \([-a, a]\).
For the current problem:
These characteristics are useful when evaluating integrals, especially definite integrals over symmetric limits like \([-a, a]\).
For the current problem:
- \( \sinh(u) \) is an odd function, so \( \sinh(-u) = -\sinh(u) \).
- Therefore, when integrating \( \cosh(u) \) from \(-1\) to \(1\), using the property of odd functions simplifies the calculation significantly.
